Algorithm
Problem Name: 563. Binary Tree Tilt
Given the root of a binary tree, return the sum of every tree node's tilt.
The tilt of a tree node is the absolute difference between the sum of all left subtree node values and all right subtree node values. If a node does not have a left child, then the sum of the left subtree node values is treated as 0. The rule is similar if the node does not have a right child.
Example 1:
Input: root = [1,2,3] Output: 1 Explanation: Tilt of node 2 : |0-0| = 0 (no children) Tilt of node 3 : |0-0| = 0 (no children) Tilt of node 1 : |2-3| = 1 (left subtree is just left child, so sum is 2; right subtree is just right child, so sum is 3) Sum of every tilt : 0 + 0 + 1 = 1
Example 2:
Input: root = [4,2,9,3,5,null,7] Output: 15 Explanation: Tilt of node 3 : |0-0| = 0 (no children) Tilt of node 5 : |0-0| = 0 (no children) Tilt of node 7 : |0-0| = 0 (no children) Tilt of node 2 : |3-5| = 2 (left subtree is just left child, so sum is 3; right subtree is just right child, so sum is 5) Tilt of node 9 : |0-7| = 7 (no left child, so sum is 0; right subtree is just right child, so sum is 7) Tilt of node 4 : |(3+5+2)-(9+7)| = |10-16| = 6 (left subtree values are 3, 5, and 2, which sums to 10; right subtree values are 9 and 7, which sums to 16) Sum of every tilt : 0 + 0 + 0 + 2 + 7 + 6 = 15
Example 3:
Input: root = [21,7,14,1,1,2,2,3,3] Output: 9
Constraints:
- The number of nodes in the tree is in the range
[0, 104].
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int findTilt(TreeNode root) {
int[] total = {0};
helper(root, total);
return total[0];
}
private int helper(TreeNode root, int[] total) {
if (root == null) {
return 0;
}
int left = helper(root.left, total);
int right = helper(root.right, total);
total[0] += Math.abs(left - right);
return left + right + root.val;
}
}
Input
Output
#2 Code Example with Javascript Programming
Code -
Javascript Programming
const findTilt = function(root) {
const tilt = { val: 0 }
dfs(root, tilt)
function dfs(root, tilt) {
if (!root) return 0
let left = dfs(root.left, tilt)
let right = dfs(root.right, tilt)
tilt.val += Math.abs(left - right)
return root.val + left + right
}
return tilt.val
}
Input
Output
#3 Code Example with Python Programming
Code -
Python Programming
class Solution:
def findTilt(self, root):
"""
:type root: TreeNode
:rtype: int
"""
def traverse(node):
if not node: return 0
left = traverse(node.left)
right = traverse(node.right)
res.append(abs(right -left))
return node.val + left + right
res = []
traverse(root)
return sum(res)
Input
Output
#4 Code Example with C# Programming
Code -
C# Programming
using System;
namespace LeetCode
{
public class _0563_BinaryTreeTilt
{
public int FindTilt(TreeNode root)
{
(_, int tilt) = ComputeSum(root);
return tilt;
}
private (int sum, int tilt) ComputeSum(TreeNode node)
{
if (node == null) return (0, 0);
(var leftSum, var leftTilt) = ComputeSum(node.left);
(var rightSum, var rightTilt) = ComputeSum(node.right);
var tile = Math.Abs(leftSum - rightSum);
return (leftSum + rightSum + node.val, tile + leftTilt + rightTilt);
}
}
}
Input
Output